Problem: $ B = \left[\begin{array}{rrr}4 & 2 & 3 \\ 1 & 1 & 4\end{array}\right]$ $ D = \left[\begin{array}{rr}-1 & 4 \\ 5 & -1 \\ 3 & 1\end{array}\right]$ What is $ B D$ ?
Answer: Because $ B$ has dimensions $(2\times3)$ and $ D$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B D = \left[\begin{array}{rrr}{4} & {2} & {3} \\ {1} & {1} & {4}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{4} \\ {5} & \color{#DF0030}{-1} \\ {3} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{-1}+{2}\cdot{5}+{3}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-1}+{2}\cdot{5}+{3}\cdot{3} & ? \\ {1}\cdot{-1}+{1}\cdot{5}+{4}\cdot{3} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-1}+{2}\cdot{5}+{3}\cdot{3} & {4}\cdot\color{#DF0030}{4}+{2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{1} \\ {1}\cdot{-1}+{1}\cdot{5}+{4}\cdot{3} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{-1}+{2}\cdot{5}+{3}\cdot{3} & {4}\cdot\color{#DF0030}{4}+{2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{1} \\ {1}\cdot{-1}+{1}\cdot{5}+{4}\cdot{3} & {1}\cdot\color{#DF0030}{4}+{1}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}15 & 17 \\ 16 & 7\end{array}\right] $